Question

50 ml of mixture of NH3 AND H2 are decomposed to yield N2 AND H2 AND to the resulting mixture 40ml of O2 is added and the mixture is sparked to yield H2O. when the resulting mixture of gases were passed through alkaline pyrogallol 6ml contraction was observed. calculate the percentage composition of NH3 in the original mixture

Answer 1

Let the volume of Ammonia be  V ml  in the mixture.  So Hydrogen gas has the volume of 50 - V  ml.

2 NH3 (g) ==> 2 N2 (g) + 3H2 (g)

So after decomposition of Ammonia, we get V ml of Nitrogen and V*3/2  ml  of Hydrogen.  
Total volume of H2 = (50 - V) + V *3/2  = 50 + V/2   ml.
     This volume is less than  75 ml  as V < 50 ml

Now 40 ml of O2 is added.  The amount of water formed and oxygen consumed is obtained as:

    2 H2 + O2 ==> 2 H2O
So 50+ V/2  ml  of Hydrogen reacts with  (25 + V/4) ml of Oxygen to form water.   The amount of Oxygen left = 40 ml - (25 +V/4) = 15 - V/4 ml


When the mixture is passed through the Alkaline Pyrogallol (C6 H3 (OH)3 ) the entire content of Oxygen is absorbed by it.  The reduction in volume is equal to that of O2.

Hence,   15 - V/4 = 6 ml    => V = 36 ml

Ammonia volume (initially) = 36 ml
Percentage = 36/50 * 100 = 72%

Answer 2

Answer: The percentage composition of NH₃ in the original mixture is 72%.

Explanation:

Let the volume of Ammonia be V ml  in the mixture thus the Hydrogen gas has the volume of 50 - V  ml.

2NH₃ (g) ⇒2 N₂ (g) + 3H₂ (g)

So after the decomposition of Ammonia, we get V ml of Nitrogen and V*3/2  ml  of Hydrogen.  

Total volume of H₂ = (50 - V) + V *3/2  = 50 + V/2   ml.

This volume is less than 75 ml since V < 50 ml

Now 40 ml of O₂ is added and the amount of water formed and oxygen consumed is obtained in the equation as:

2 H₂ + O₂⇒ 2 H₂O

So 50+ V/2  ml  of Hydrogen reacts with  (25 + V/4) ml of Oxygen to form water.The amount of Oxygen left = 40 ml - (25 +V/4) = 15 - V/4 ml

When the mixture is passed through the Alkaline Pyrogallol (C₆ H₃ (OH)₃) the entire content of Oxygen is absorbed by it.  The reduction in volume is equal to that of O₂.

Hence,   15 - V/4 = 6 ml    

⇒ V = 36 ml

Ammonia volume (initially) = 36 ml

Percentage = 36/50 * 100 = 72%

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